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As mentioned in the title, I would like to know a proof of the "well known" fact that the A6 preprojective algebra is of wild representation type.

Ideally, I would like to see an explicit two-parameter family of indecomposable modules.

If you happen to prefer type D to type A and want to give an answer there, that would be appreciated too.

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Inspired by the reference Julian gave in his comment, here's an explicit example of a two-parameter family of indecomposable representations.

First, I'll describe a one-parameter family of indecomposable representations for the preprojective algebra of type $A_5$.

Let $k$ be the coefficient field, and $\alpha\in k$. Then let $M(\alpha)$ be the representation $$\begin{array}{ccccccccc} &\begin{pmatrix}0\\1\end{pmatrix}&&\begin{pmatrix}1&0\\0&0\end{pmatrix} &&\begin{pmatrix}0&0\\1&\alpha\end{pmatrix} &&\begin{pmatrix}1&0\end{pmatrix}\\ k&\rightleftarrows&k^2&\rightleftarrows&k^2&\rightleftarrows&k^2& \rightleftarrows&k\\ &\begin{pmatrix}1&0\end{pmatrix}&&\begin{pmatrix}0&0\\1&1\end{pmatrix} &&\begin{pmatrix}0&0\\1&0\end{pmatrix} &&\begin{pmatrix}0\\\alpha\end{pmatrix}\\ \end{array}$$

It's a straightforward calculation to check that $M(\alpha)$ is indecomposable and that if $\alpha\neq\beta$ then any homomorphism $M(\alpha)\to M(\beta)$ is zero at the right hand vertex, so that a direct sum decomposition of $M(\alpha)\oplus M(\beta)$ into two non-zero summands is uniquely determined at the right hand vertex.

Now extend $M(\alpha)\oplus M(\beta)$ to a representation $N\left(\{\alpha,\beta\}\right)$ of the preprojective algebra of type $A_6$ by adding a vertex on the right hand end: $$\begin{array}{ccccccccccc} &&&&&&&&&\begin{pmatrix}0&0\end{pmatrix}\\ k\oplus k&\rightleftarrows&k^2\oplus k^2&\rightleftarrows&k^2\oplus k^2&\rightleftarrows&k^2\oplus k^2& \rightleftarrows&k\oplus k&\rightleftarrows&k\\ &&&&&&&&&\begin{pmatrix}1\\1\end{pmatrix}\\ \end{array}$$

Then it follows easily from the properties stated for the representations $M(\alpha)$ that $N\left(\{\alpha,\beta\}\right)$ is indecomposable and $N\left(\{\alpha,\beta\}\right)\not\cong N\left(\{\alpha',\beta'\}\right)$ unless $\{\alpha,\beta\}=\{\alpha',\beta'\}$.

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    $\begingroup$ Nice. The statement that there are no Homs between M(alpha) and M(beta) is false, but irrelevant for proving the desired properties about N({alpha,beta}). $\endgroup$
    – Peter McNamara
    Apr 10, 2015 at 15:26
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    $\begingroup$ @PeterMcNamara Ah, you're right. But there are no Homs whose right hand component is non-zero, which is what is important to deduce that $N\left(\{\alpha,\beta\}\right)$ is indecomposable. I'll edit to correct this. $\endgroup$
    – Jeremy Rickard
    Apr 10, 2015 at 15:45
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    $\begingroup$ A similar construction also works in type D_5. You have a one parameter family for D_4 in dimension (1,1,2,1) (verticies numbered so that the third one is the central one). So Jeremy's construction gives a two-parameter family in dimension (2,2,4,2,1). $\endgroup$
    – Peter McNamara
    Apr 15, 2015 at 7:10

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